I read in Milnor's article "Link groups", where he defines invariants to classify links up to link homotopy, that the linking number is a complete invariant which can tell almost trivial two components links apart up to link homotopy. My question is how do I prove that the linking number is a complete invariant for link homotopy?
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The linking number is the same as the homology class that one component represents in the complement of the other. You can reduce any $2$-component link to a normal form by first homotoping one component to be an unknot. Then the complement of this component is homeomorphic to a solid torus, which has $\pi_1\cong\mathbb Z$. So all loops can be homotoped to one representative for each $n\in\mathbb Z$, where $n$ is the linking number.
answered May 20, 2017 at 15:17
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$\begingroup$ Thank you for your answer! It helped me to look at the problem differently. To be more precise, I'd like to prove that if the linking number of the two components of the link is zero then the 2-link is trivial (up to link homotopy). Can you provide more help on this? $\endgroup$– SukiCommented May 21, 2017 at 14:35
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$\begingroup$ @SoukainaHamri when you say 2-link, are you referring to a 2-component link or a link which has 2-dimensional components? I.e. a link of spheres in 4-space? If you are just referring to a link with 2 components, then the case of linking number 0 is a special case of my argument above. $\endgroup$ Commented May 21, 2017 at 22:26
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$\begingroup$ I meant the first case. Thank you for you help. $\endgroup$– SukiCommented May 21, 2017 at 23:11
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$\begingroup$ So just to clarify: do crossing changes on one component until it is unknotted. Then the other component is null homotopic in its complement, since it is a closed loop representing a trivial homology class in a solid torus. $\endgroup$ Commented May 21, 2017 at 23:48
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$\begingroup$ Thanks. I spent some time trying to fully understand your argument, now I'll try to prove that a curve that is homologically trivial in a solid torus is null-homotopic. $\endgroup$– SukiCommented May 23, 2017 at 21:59